## Four Clocks Problem

 Quote by dingpud I am going to take a crack at this: Assign arbitrary subscripts to each ball as the following: W1 W2 R1 R2 B1 B2 Step 1: Place B1+B2+R1 on the first scale. Place W1+W2+R2 on the second scales. Whichever scales reads the higher value contains the heavy red ball. Record the high weight measurement and record as malpha Label the heavy ball as Rheavy Label the light ball as Rlight (Here is where a possible technicallity may occur) Step 2: Use only 1 scale for this measurement: Place on the scale B1+W1+Rheavy If the reading on the scale is greater than malpha then B1 and W1 are both heavy while B2 and W2 are both light. If the reading on the scale is less than malpha then B1 and W1 are both light while B2 and W2 are both heavy. If the reading on the scales is equal to malpha then use the fourth scale with the same set-up as the third weigh in, however, swap out B1 with B2. Apply the same logic rules as above, and you should be able to differentiate between the heavy and light balls. Let me know what you think.... I wanted to give the answer in pure mathematical / algebraic form, but don't know when I'll get time to attempt that.
Please check . I am pasting ur solution.
Step 2:
Use only 1 scale for this measurement:
Place on the scale B1+W1+Rheavy
If the reading on the scale is greater than malpha then B1 and W1 are both heavy while B2 and W2 are both light.

There are no ways to measure absolute value of malpa as suggested by you. Only six balls and a balance with two pans. No weights are available.

 Quote by dingpud Has anyone been able to verify my take at weighing the Christmas balls?

Please check . I am pasting ur solution.
Step 2:
Use only 1 scale for this measurement:
Place on the scale B1+W1+Rheavy
If the reading on the scale is greater than malpha then B1 and W1 are both heavy while B2 and W2 are both light.
.............
Now
There are no ways to measure absolute value of malpa as suggested by you. Only six balls and a balance with two pans. No weights are available.
 OK, so it is a balance scale, and not necessarily a weighing scale. Alright, I'll have to look at it again then....

 Quote by dingpud OK, so it is a balance scale, and not necessarily a weighing scale. Alright, I'll have to look at it again then....
Yes, Sir.
There is only one balance scale with two pans. No weights available.
You can use scale twice only.
Please excuse me if language of my question created any confusion.
 Language, not a problem.... Mathematics is the universal language, right?

 Quote by dipinsingh I think it is not correct. Pls check the statement. The third reads 3:45 and gains 1hour and 20 minutes a day. I think u have solved with condition of loosing time 1 hr 20 min. Pls check. Excuse me if i m wrong.
You are absolutely correct. :)

Btw, here's my take on the christmas tree problem...
Let the blue, red and white balls be labelled A, B, C, respectively. Weigh A1+B1 against A2+C1. If A1+B1 == A2+C1, weigh A1 versus A2. If A1 > A2, then the heavier balls are A1, B2 and C1; otherwise, it is A2, B1 and C2. If A1+B1 > A2+C1, weigh A1+A2 against B1+C1 next. If A1+A2 > B1+C1, then the heavier balls are A1, B2, C2; if A1+A2 < B1+C1, the heavier balls are A1, B1, C1; otherwise, if A1+A2 == B1+C1, the heavier balls are A1, B1, C2. If, in the first step, A1+B1 < A2+C1, weigh A1+A2 against B1+C1 next. If A1+A2 > B1+C1, then the heavier balls are A2, B2, C2; if A1+A2 < B1+C1, the heavier balls are A2, B1, C1; otherwise, if A1+A2 == B1+C1, the heavier balls are A2, B2, C1.

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